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          Problem 460
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<h1 id="Problem-460"><a href="#Problem-460" class="headerlink" title="Problem 460"></a><a href="https://projecteuler.net/problem=460" target="_blank" rel="noopener">Problem 460</a></h1><hr>
<p><strong>An ant on the move</strong></p>
<p>On the Euclidean plane, an ant travels from point A(0, 1) to point B(d, 1) for an integer d.</p>
<p>In each step, the ant at point (x<sub>0</sub>, y<sub>0</sub>) chooses one of the lattice points (x<sub>1</sub>, y<sub>1</sub>) which satisfy x<sub>1</sub> ≥ 0 and y<sub>1</sub> ≥ 1 and goes straight to (x<sub>1</sub>, y<sub>1</sub>) at a constant velocity v. The value of v depends on y<sub>0</sub> and y<sub>1</sub> as follows:</p>
<ul>
<li>If y<sub>0</sub> = y<sub>1</sub>, the value of v equals y<sub>0</sub>.</li>
<li>If y<sub>0</sub> ≠ y<sub>1</sub>, the value of v equals (y<sub>1</sub> - y<sub>0</sub>) / (ln(y<sub>1</sub>) - ln(y<sub>0</sub>)).</li>
</ul>
<p>The left image is one of the possible paths for d = 4. First the ant goes from A(0, 1) to P<sub>1</sub>(1, 3) at velocity (3 - 1) / (ln(3) - ln(1)) ≈ 1.8205. Then the required time is sqrt(5) / 1.8205 ≈ 1.2283.<br>From P<sub>1</sub>(1, 3) to P<sub>2</sub>(3, 3) the ant travels at velocity 3 so the required time is 2 / 3 ≈ 0.6667. From P<sub>2</sub>(3, 3) to B(4, 1) the ant travels at velocity (1 - 3) / (ln(1) - ln(3)) ≈ 1.8205 so the required time is sqrt(5) / 1.8205 ≈ 1.2283.<br>Thus the total required time is 1.2283 + 0.6667 + 1.2283 = 3.1233.</p>
<p>The right image is another path. The total required time is calculated as 0.98026 + 1 + 0.98026 = 2.96052. It can be shown that this is the quickest path for d = 4.</p>
<center><img src="https://projecteuler.net/project/images/p460_ant.jpg"></center>

<p>Let F(d) be the total required time if the ant chooses the quickest path. For example, F(4) ≈ 2.960516287.<br>We can verify that F(10) ≈ 4.668187834 and F(100) ≈ 9.217221972.</p>
<p>Find F(10000). Give your answer rounded to nine decimal places.</p>
<hr>
<p><strong>蚂蚁旅行</strong></p>
<p>在欧几里德平面上，一只蚂蚁打算从点A(0, 1)运动到点B(d, 1)，其中d是一个整数。</p>
<p>每一步，位于点(x<sub>0</sub>, y<sub>0</sub>)的蚂蚁会选择一个格点(x<sub>1</sub>, y<sub>1</sub>)，满足x<sub>1</sub> ≥ 0及y<sub>1</sub> ≥ 1，然后以恒定速率v径直移动过去。v的值取决于y<sub>0</sub>和y<sub>1</sub>，如下所示：</p>
<ul>
<li>若y<sub>0</sub> = y<sub>1</sub>，则v的值为y<sub>0</sub>。</li>
<li>若y<sub>0</sub> ≠ y<sub>1</sub>，则v的值为(y<sub>1</sub> - y<sub>0</sub>) / (ln(y<sub>1</sub>) - ln(y<sub>0</sub>))。</li>
</ul>
<p>左图是当d = 4时的其中一条路径。首先蚂蚁从点A(0, 1)移动到点P<sub>1</sub>(1, 3)，速率为(3 - 1) / (ln(3) - ln(1)) ≈ 1.8205，所需时间为sqrt(5) / 1.8205 ≈ 1.2283。<br>再从点P<sub>1</sub>(1, 3)移动到点P<sub>2</sub>(3, 3)，蚂蚁的速率为3，所需时间为2 / 3 ≈ 0.6667。从点P<sub>2</sub>(3, 3)移动到点B(4, 1)，蚂蚁的速率是(1 - 3) / (ln(1) - ln(3)) ≈ 1.8205，所需时间为sqrt(5) / 1.8205 ≈ 1.2283。<br>因此，总耗时为1.2283 + 0.6667 + 1.2283 = 3.1233。</p>
<p>右图是另一条可行路径，总耗时为0.98026 + 1 + 0.98026 = 2.96052，可以看出这是d = 4时的最快路径。</p>
<center><img src="https://projecteuler.net/project/images/p460_ant.jpg"></center>

<p>记F(d)是蚂蚁选择最快路径时所需的耗时，例如F(4) ≈ 2.960516287。<br>我们可以验证F(10) ≈ 4.668187834以及F(100) ≈ 9.217221972。</p>
<p>求F(10000)，保留9位小数。</p>
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